# Height of Ball

• Dec 14th 2008, 10:00 PM
magentarita
Height of Ball
The formula for the height of a ball as a function of time is given by the equation h = -16t^2 + vt + h, where h is the height of a ball in feet, v is the initial velocity of the ball in feet per second, h is the initial height of the ball in feet, and t is the time in seconds after the ball was thrown.
If a ball is thrown from an initial height of 5 feet at an initial velocity of 20 feet per second, what is its height after 1 second?

Do I plug and chug here?

• Dec 14th 2008, 10:13 PM
tester85
Quote:

Originally Posted by magentarita
The formula for the height of a ball as a function of time is given by the equation h = -16t^2 + vt + h, where h is the height of a ball in feet, v is the initial velocity of the ball in feet per second, h is the initial height of the ball in feet, and t is the time in seconds after the ball was thrown.
If a ball is thrown from an initial height of 5 feet at an initial velocity of 20 feet per second, what is its height after 1 second?

Do I plug and chug here?

Yes that's right but remember to plug in h which is the initial height into the correct h in the equation as the equation has 2 h's in it.
• Dec 15th 2008, 09:33 PM
magentarita
ok...
Quote:

Originally Posted by tester85
Yes that's right but remember to plug in h which is the initial height into the correct h in the equation as the equation has 2 h's in it.

So, h = 5 in the function.

Plug the rest for v and t and simplify, right?
• Dec 15th 2008, 09:36 PM
Chris L T521
Quote:

Originally Posted by magentarita
So, h = 5 in the function.

Plug the rest for v and t and simplify, right?

Correct. The equation you should get should be $h\left(t\right)=-16t^2+20t+5$

Now, what do you think $h\left(1\right)$ should be?